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1). 02). 0,13). 3,2

1) 02)0,13)3,2hope it will help u

Factorising it by middle term splitting :-

x² + 4x + 3

x² + 3x + x + 3

x ( x + 3 ) + 1 ( x + 3 )

( x + 3 ) ( x + 1 )

• ( x + 3 ) = 0

x = ( - 3 )

• ( x + 1 ) = 0

x = ( - 1 )alpha=-3 and beta=-1zeroes of new equation are1+a/b and 1+b/a1+a/b=1-3/-1=2and 1+b/a=1-1/-3=0So, the Zeros of new equation are 2 and 0

• Sum of the Zeros are :-

0 + 2 = 2

• Product of the Zeros are :-

0 × 2 = 0

♯ To form the quadratic equation we have formula as :-

x² - ( sum of Zeros )x + (product of Zeros)

Putting value in it !!

x² - 2x + 0

So, the required quadratic equation isx² - 2x .

12/13 + x =0(12/13)+(-12/13)= 0hence,provedas+ of - equals -so here(12/13)-(12/13)=0example:- a-a= 0

-12/13 is correct answer of following question

cos 58° = cos(90 - 32°) = sin 32°sin 22° = sin(90 - 68°) = cos 68° cos 38° = cos(90 - 52°) = sin 52°tan 18° = cot 72°tan 35° = cot 55°[cos 58°/sin 32°] + [sin 22°/cos 68°] - [(cos 38° cosec 52°)/(tan 18° tan 35° tan 60° tan 72° tan 55°)] = [sin 32°/sin 32°] + [cos 68°/cos 68°] - [(sin 52°/sin 52°)/(cot 72° cot 55° tan 60° tan 72° tan 55°)] = [1] + [1] - [1/√3] (∵ tan 60° = √3) = (2√3 - 1)/√3 = (6 - √3)/3 (∵ Rationalise the denominator)

⇒(cos(90-32)/sin32)+(sin(90-68)/cos68)-cos(90-52)cosec52)/tan(90-72).tan35. tan60.tan72.tan(90-35)

⇒(sin32/sin32)+(cos68/cos68)-sin52.cosec52/cot72.tan35.tan60.tan72.cot35

⇒1+1-1/1*1*tan60

⇒2-1/√3

⇒(2√3-1)/√3

by multiply nominator and denominator by √3

⇒(6-√3)/3

Given : Pⁿ is linear space of polynomial of degree n or less.

To generate Pⁿ we need

{ 1, x, x², .... xⁿ}

Therefore, dim Pⁿ = n + 1

Tq

Let the two digit number be xConditions:x is multiple of 11divisible by 7q+4=1577 is the number which satisfy all the above condition77 is multiple of 11when divided by 7 leaves no remainder an gives quotient as 11when 11+4=15Thus x=77Like if you find it useful

d

b

a

c

please read the question

again and give answer

_13/12+x=79/36 x=79/36+13/12=118/36=59/18

Answer is -5/2

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Assuming that you meancos 58/sin 32 + sin 22/cos 68- (cos 38 csc 52)/(tan 18*tan 35*tan 60*tan 72*tan 55)

Use the cofunction identities (in degrees):sin(90 - t) = cos t, cos(90 - t) = sin t, and tan(90 - t) = cot t.

This simplifies the expression in question tosin(90 - 58)/sin 32 + cos(90 - 22)/cos 68- (cos 38 csc 52)/(tan 18*tan 35*tan 60*cot(90-72)*cot(90-55))

= 1 + 1 - (cos 38 csc 52)/(tan 18*tan 35*tan 60*cot 18*cot 35)

Now, use csc t = 1/sin t and cot t = 1/tan t:2 - (cos 38/sin 52)/(tan 18*tan 35*tan 60*(1/tan 18)*(1/tan 35))= 2 - (cos 38/sin 52)/tan 60= 2 - (cos 38/cos(90-52))/tan 60, via cofunction identity= 2 - 1/tan 60= 2 - 1/√3.

For any quadratic polynomial ax²+bx+c the zeroes are precisely the x- coordinates of the points where the graph of y = ax² + bx +c intersects the X- axis.

It is Clear from the graph that curve intersect the x-axis at 3 points. Hence the number of zeros of the polynomial y = p(x) are three.

A pair of irrational number whose product is a rational number is (a) √16√4√16 x √4= √64= 8

next

let x= √6+√6+√6...

Squaring both sides, we get

x2=6+√6+√6+√6...

x2=6+x

x2-x-6=0

x2-3x+2x-6=0

x(x-3)+2(x-3)=0

(x-3)(x+2)=0

x=-2,3

As √6+√6+√6... is positive

So, the value of √6+√6+√6... is3

5t-3t=+3-52t=-2t=-2/2t =-1

5t-3t=3-52t=-2t=2/2t=-1

5t-3t=+3-53t=-2t=2/2t=1

[ b/4 + c/3]²

(b/4)² + (c/3)² + 2 × (b/4) × (c/3)

b²/4 + c²/9 + bc/6

iii) 36 = 9x -> x = 36/9 = 4 iv) 5t = -20 -> t = -20/5 = -4 ix) -2x = 1 -> x = -1/2

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(21x*x*x*x-14x*x+7x)÷(7x*x*x) = 7x(3x*x*x-2x+1)÷(7x*x*x) = 3x- 2/x + 1/(x*x)

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1as 1*2=21*3=3any number multiplied by 1 will give the number itself

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A set of Rational numbers are always closed under all of the operations of subtraction, because doing subtraction always results in p/q form

Given factor: x² + 2x + k = 0

Given polynomial: 2x⁴ + x³ -14x² + 5x + 6

Divide the polynomial by the factor

x²+ 2x + k ) 2x⁴ + x3 -14x²+ 5x + 6 ( 2x² - 3x +(- 8 - 2k)

2x⁴ + 4x³ +2kx² ( substract)

------------------------------

- 3x³ +(-14 - 2k)x² + 5x

- 3x³ - 6x² - 3kx ( substract)

------------------------------

(- 8 - 2k) x² +( 5 + 3k)x + 6

(- 8 - 2k) x² +(-16 - 4k)x + (- 8k - 2k²) ( substract)

-----------------------------------------------------------------

( 21 + 7k)x + (6 + 8k + 2k²)

The remainder is: ( 21 + 7k)x + (6 + 8k + 2k²) = 0

21 + 7k = 0 ⇒ k = -3.

The factors are x² + 2x - 3 = 0 and 2x² - 3x - 2 = 0

x² + 3x - x - 3 = 0 and 2x² - 4x + x - 2 = 0

x( x + 3 )-1( x + 3) = 0 and 2x (x - 2) + 1(x - 2) = 0

(x - 1)( x + 3) = 0 and (2x + 1)(x - 2) = 0

x = 1 ,3 ,-1 / 2 and 2.

The zeros are 1 ,3 ,-1 / 2 and 2.

For E1= 1-0.6=0.4 For E2= 1-0.4=0.6

2x²+7x+3 = 0

2x²+6x+x+3 = 0

2x(x+3)+1(x+3) = 0

(2x+1)(x+3) = 0

x= -1/2, -3

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